It looks like a case for Watson’s lemma with f(t) = e −1 t . However: e −1 t and all its derivatives vanish at t = 0 ⇒ Watson’s lemma would yield I(x) = 0 The dominant contribution to the integral comes from the maximum of the integrand. In this case the maximum depends on x, Note: tmax = 1 √ x • a maximum that depends on x is called a movable maximum. • to apply Laplace’s method to a movable maximum transform t into a coordinate in which the maximum is fixed, independent of x. We could introduce a shift of the coordinate s = t − tmax This would lead to complicated expressions in e −1 t . Instead, the maximum can be fixed by a rescaling s = √ xt smax = 1 I(x) = 1 � ∞ √ x 0 e −√ x( 1 s +s) ds Using (26) and noting that √ x plays the role of x in (26) we get and φ ′′ (1) = −2 Note: 3 I(x) ∼ 1 � √ x 2π 2 √ x e−2√ x = √ π x 3 4 e −2√ x • I(x) decreases faster than any power. Watson’s lemma can only generate results that depend on x as a power series. 3 Striling’s formula in Hinch p.33 is nice example for movable maximum and for important formula 57